Question

Assume that the heights of females in a certain village are normally distributed with a mean of 69 in. and standard deviation of 33 in. What percent of these females are between 6060 in. and 78 in.​ tall?

Answer 1

Answer: 21.28%

Explanation:

Given : Mean height of females =
`\mu=69\text{ in.}`

Standard deviation :
`\sigma=33\text{ in.}`

Let X be a random variable that represents the heights of females in village .

We assume that that the heights of females in a certain village are normally distributed .

Z-score :
`z=(x-\mu)/(\sigma)`

For x = 60 in.

`z=(60-69)/(33)\approx-0.27`

For x = 78 in.

`z=(78-69)/(33)\approx0.27`

Now by using standard normal distribution table , the probability that females are between 60 in. and 78 in.​ tall :-

`P(60<x<78)=P(-0.27<z<0.27)\\\\=P(z<0.27)-P(z<-0.27)\\\\=0.6064198-0.3935801=0.2128397\approx0.2128=21.28\%`

Hence, the percent of these females are between 60 in. and 78 in.​ tall =21.28%